In my homework I have:
$G=D_8$ (Dieder) and $H=\langle r^2 \rangle$
It it said is that $G/H \cong \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$
Why is that? Why not $\mathbb{Z}/4\mathbb{Z}$?
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6$G/N={[1],[r],[\sigma ],[r\sigma ]}$ and as you see, there are no element of order 4. – Surb Dec 28 '20 at 09:44
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Typo, have corrected now, thx – Daniel Dec 28 '20 at 13:37
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Just Calculate $G/H$: Clearly, $$G/H=\{g o H|g \in G \}$$ So, $$G/H=\{1 o <r^2>,r o <r^2>,s o <r^2>,rs o <r^2>\}$$ where $s$ is reflection and $r$ is rotation and $o$ is composition operation of $G$. Clearly $G/H$ is not cyclic as it does not have any generator but $\mathbb{Z_4}$ is cyclic.
Vats Y
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1Why do you have addition between $r+r^2$,shouldn't it be multiplication? – Daniel Dec 28 '20 at 13:59
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I still don't see why it is addition. Both G and H are multiplicative groups. Do you mean the composition "$\cdot$" instead of addition "+"? – Daniel Dec 28 '20 at 14:55
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Please accept the answer if it is clear or more thing I have to add on? – Vats Y Dec 28 '20 at 15:55
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Since $Z(D_8)$ is generated by $r^2$ we have $G/H=D_8/Z(D_8)$. If it would be cyclic then $D_8$ would be abelian, a contradiction. Hence $D_8/Z(D_8)$ must be isomorphic to $C_2\times C_2$.
Reference: If $G/Z(G)$ is cyclic, then $G$ is abelian
Dietrich Burde
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